Multiply the following complex numbers: $({-4+4i}) \cdot ({3+2i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-4+4i}) \cdot ({3+2i}) = $ $ ({-4} \cdot {3}) + ({-4} \cdot {2}i) + ({4}i \cdot {3}) + ({4}i \cdot {2}i) $ Then simplify the terms: $ (-12) + (-8i) + (12i) + (8 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -12 + (-8 + 12)i + 8i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -12 + (-8 + 12)i - 8 $ The result is simplified: $ (-12 - 8) + (4i) = -20+4i $